Question

A uniform rod of mass $m$ and length $l$ can rotate in a vertical plane about a smooth horizontal axis hinged at point $H$. Find the force exerted by the hinge just after the rod is released as shown in the figure.

• A. $mg\sqrt{7}$
• B. $\frac{mg\sqrt{10}}{5}$
• C. $\frac{mg}{5}$
• D. $\frac{mg}{\sqrt{5}}$

Answer 1

The correct option is B $\frac{mg\sqrt{10}}{5}$


Taking torque equation about hinge $H$,
${\tau }_{\text{net}}=I\alpha$
$\Rightarrow mg\cos {37}^{\circ }\times \frac{l}{2}=\frac{m{l}^2}{3}\alpha$
$\Rightarrow \alpha =\frac{6g}{5l}$
[torque due to hinge reaction is zero]

Let the hinge force components be $N_1$ and $N_2$
Tangential acceleration of COM of the system
$a_t=\frac{l}{2}\alpha =\frac{3g}{5}$

Writing force equation perpendicular to the rod,
$mg\cos {37}^{\circ }-N_1=m{a}_t$
$\Rightarrow mg\times \frac{4}{5}-N_1=m\times \frac{3g}{5}$
$\Rightarrow N_1=\frac{mg}{5}$
Since, angular velocity $\left(\omega \right)=0$ at the time of release and net force along the rod is zero since there is no acceleration towards the center,
$N_2=mg\sin {37}^{\circ }=\frac{3mg}{5}$

Therefore, hinge force $\left(N\right)$ is
$N=\sqrt{N_1^2+N_2^2}=\sqrt{{\left(\frac{mg}{5}\right)}^2+{\left(\frac{3mg}{5}\right)}^2}=\frac{mg\sqrt{10}}{5}$
Hence, option (b) is the correct answer.